We make extensive use of limiting reagents in this lab. This question sets the s

Question

We make extensive use of limiting reagents in this lab. This question sets the scene and also provides practice for the sample calculations, which you will include throughout the report. According to the instructions, 1.5 g of KI is added to a solution containing 0.60 g of K_3Fe(CN)_6, and the following reaction occurs: 2 Fe(CN)_6^3- + 2 r rightarrow 2 Fe(CN)_6^4- + I2. Show, by calculation, that KI is, indeed, in excess (as stated on Page 29). Calculate the number of moles of I_2 formed from these amounts of reagents. In each calculation, show the complete setup, including units and consistent significant figures.

Explanation / Answer

Answer – We are given the mass of KI = 1.5 g , mass of K3Fe(CN)6 = 0.60 g

Reaction –

2 Fe(CN)63- + 2 I- -----> 2 Fe(CN)64- + I2

First we need to calculate the moles of each given

Moles of KI = 1.5 g / 165.998 g.mol-1 = 0.00904 moles

Moles of K3Fe(CN)6 = 0.60 g/ 329.24 g.mol-1 = 0.00182 moles

We know, 1 moles of KI = 1 moles of I-

So, moles of I- = 0.00904 moles

Moles of K3Fe(CN)6 = moles of Fe(CN)63- = 0.00182 moles

a) Now from the balanced equation we know the mole ratio between both reactant and Fe(CN)64- is 1:1

So, moles of Fe(CN)64- from the I- = 0.00904 moles

moles of Fe(CN)64- from the Fe(CN)63- = 0.00182 moles

so the moles of Fe(CN)64- = 0.00182 moles is lowest and it is got from the Fe(CN)63- so the K3Fe(CN)6 is the limiting reactant and the KI is in the excess form.

b) We already calculated the moles of both reactant and from that moles of I2 as follow –

First form the I-

from the balanced reaction

2 moles of I- = 1 moles of I2

So, 0.00904 moles of I- = ?

= 0.0045 moles of I2

First form the Fe(CN)63-

from the balanced reaction

2 moles of Fe(CN)63-= 1 moles of I2

So, 0.00182 moles of Fe(CN)63-= ?

= 0.00091 moles of I2

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